Question

If |z + 1| = z + 2(1 + i), then find z.

Answer 1

Given that: |z + 1| = z + 2(1 + i)

Let z = x + iy

So, |x + iy + 1| = (x + iy) + 2(1 + i)

⇒ |(x + 1) + iy| = x + iy + 2 + 2i

⇒ |(x + 1) + iy| = (x + 2) + (y + 2)i

⇒ `sqrt((x + 1)^2 + y^2)` = (x + 2) + (y + 2)i   ......`[because |x + iy| = sqrt(x^2 + y^2)]`

Squaring both sides, we get,

(x + 1)² + y² = (x + 2)² + (y + 2)² .i² + 2(x + 2)(y + 2)i

⇒ x² + 1 + 2x + y² = x² + 4 + 4x – y² – 4y – 4 + 2(x + 2)(y + 2)i

Comparing the real and imaginary parts, we get

x² + 1 + 2x + y² = x² + 4x – y² – 4y and 2(x + 2)(y + 2) = 0

⇒ 2y² – 2x + 4y + 1 = 0   ......(i)

And (x + 2)(y + 2) = 0  .....(ii)

x + 2 = 0 or y + 2 = 0

∴ x = –2 or y = –2

Now put x = –2 in equation (i).

2y² – 2 × (–2) + 4y + 1 = 0

⇒ 2y² + 4 + 4y + 1 = 0 

⇒ y² + 4y + 5 = 0

b² – 4ac = (4)² – 4 × 2 × 5

16 – 40 = –24 < 0 no real roots.

Put y = –2 in equation (i).

2(–2)² – 2x + 4(–2) + 1 = 0

8 – 2x – 8 + 1 = 0

⇒ x = `1/2` and y = –2

Hence, z = x + iy = `(1/2 - 2i)`.